How do you graph or define a^x when a is negative and x is not an integer?

I plugged this in wolfram alpha and got 2 graphs (one real and one imaginary). I know if you have x = 0.5 it's going to be imaginary; what if you do x = 0.3? My calculator wouldn't register that; is that something even calculable/graphable?

Otherwise, ax can be defined using basic properties of exponents for rational, non integer x and for real, irrational x by using a sequence of rational numbers which converges to x. This only works when x is real and a is not negative.

to as many terms as you would like. I will use 20 terms (more than enough for our purposes). Keep in mind that these numbers are irrational and have infinite decimal expansions. I am only using six decimal places.

for real, irrational x by using a sequence of rational numbers which converges to x. This only works when x is real and a is not negative

Why does it not work for negative a, too? I can easily define it for x integer and for reciprocals as the (1/x)th principial root and therefore for rational x as ap/q = (a1/q)^p, right? Now I can define it for irrational x as a limit again. Where's the problem?

The response above was on-point, but somehow a bit of a smokescreen. This is perfectly well-defined unless you're trying to evaluate by taking a limit.

The problem is this: imagine that you're trying to evaluate (-a)x as a limit; you need to assume that convergence works. If you don't know precisely what a limit is, don't worry - it's what you think it ought to be. Just note that it doesn't necessarily exist - the alternating sequence of -1 and 1 doesn't have one, since it flip-flops around forever, never "settling in" on a particular value. It has some data (the sequence of 1s) that suggests a limit of 1, and some data (the -1s) suggesting a different limit of -1, so it can't have a limit.

So in particular, (-1)x for x irrational isn't going to be easy to calculate this way, since any irrational is arbitrarily close to even-topped and odd-topped fractions (still true if we make sure that they're written in lowest-terms) so we have some data that suggests a limit of +1 as before and some that suggests a limit of -1.

I'm hiding some of the story, but basically what I'm saying is that the topological explanation (in terms of limits) for how this calculation works starts to fall apart and become inconsistent (you can get several contradictory answers to one question using these methods) so we need to trade this set of ideas in for a slightly more powerful algebraic formalism that actually gives us a single answer consistent with the laws of logarithms (it's essentially what they're for!), but doesn't leave us with that warm, Disney glow that we get with a simple geometric or topological interpretation of what's happening.

As for why you're getting two graphs, I think this is what you're seeing.

What you have is a function [; f:R\to C ;] taking the real numbers to the complex numbers, but the answer you're getting is still a one-dimensional output. Essentially, it's taking the real line and "bending" it around in complex space, sort of like an earlier animation I made in 3D or that same animation in 2D.

Essentially, you have a complex-curve for your graph and it's not going to be easy to think about at first. fuccgirl1's explanation of complex exponentiation is good enough at explaining what ax looks like for a<0 and x not necessarily integer-valued. But I think you might also want a good understanding of what this thing looks like so try running this code in Mathematica.

The imaginary part of ax is periodic, so instead of being jagged or incalculable, it turns into sine waves. For example, (-2)x = 2x cos(pi x)+_i_ 2x sin(pi x). It's basically as sine wave that scales with |a|x, and a phase shifted version in the imaginary part to cancel it out with a sign change.

This, where b and c are the bounds on your interval, and a is positive of course. This can also be found in your calculus book.

If ax were undefined at every irrational point, then the function would have uncountably many discontinuities, thus making the Riemann integral impossible.If we're talking Lebesgue integral, then if the function is defined only over a set of measure 0, then the integral must be 0, which it isn't.

For [;a;] positive you can define ax for x positive (including irrational) without resorting to using logarithms. You just use that if [;q_i\in\mathbb Q;] is a Cauchy sequence of rationals then [;a^{q_i};] is a Cauchy sequence of real numbers.

Sure, for non-positive [;a;] it is little disappointing that the definition isn't something based on "first principles," but how else are you going to do it? Using the logarithm and the exponential function, you get an extension that is holomorphic away from the branch cut. This is the best one could ever hope for.

Edit: In some sense it has to be done this way. Any two holomorphic extensions of ax will have to agree on a neighborhood of [;\{\frac{p}{q} \in \mathbb Q : p \text{ is even and } q \text{ is odd }\};] since it is dense in the positive real axis. This is enough to force them to agree on a neighborhood of the positive real axis. So, it really doesn't matter how you get there, just as long as you do.

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